In triangle $ABC$, $BC = 20 \sqrt{3}$ and $\angle C = 30^\circ$.  Let the perpendicular bisector of $BC$ intersect $BC$ and $AC$ at $D$ and $E$, respectively.  Find the length of $DE$.
We have that $D$ is the midpoint of $BC$, and that $CD = BC/2 = 20 \sqrt{3}/2 = 10 \sqrt{3}$.

[asy]
unitsize(3 cm);

pair A, B, C, D, E;

A = dir(133);
B = dir(193);
C = dir(-13);
D = (B + C)/2;
E = extension(A, C, D, D + rotate(90)*(B - C));

draw(A--B--C--cycle);
draw(D--E);

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, NE);
[/asy]

Also, triangle $CED$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle, so $DE = CD/\sqrt{3} = 10 \sqrt{3}/\sqrt{3} = \boxed{10}$.